158-NCERT-10-4-Quadratic Equations - Ex-4.3

NCERT

10th Mathematics

Exercise 4.3

Topic: 4 Quadratic Equations

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EXERCISE 4.3

Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2x² – 7x + 3 = 0         (ii) 2x² + x – 4 = 0

(iii) 4x² + 4√3x + 3 = 0     (iv) 2x² + x + 4 = 0

Explanation:

1) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

ax² + bx + c = 0

x² + (b/a)x + (c/a) = 0

x² + (b/a)x = - (c/a)
x² + (b/a)x = - (c/a) ------------------ equation 1.

2) In the quadratic equation, for getting the perfect square, 

last term = [(middle term)²]/4 ------------------ equation 2.

3) From equation 1 and equation 2, last term = [(b/a)²]/4.

4) Adding [(b/a)²]/4 to both sides of equation 1, we get,

x² + (b/a)x + [(b/a)²]/4 = [(b/a)²]/4 - (c/a)

x² + (b/a)x + (b/2a)² = (b/2a)² - (c/a)
x² + (b/a)x + (b/2a)² = (b)²/4a² - (c/a)

(x + b/2a)² = (b² - 4ac)/4a²

(x + b/2a) = ± √[(b² - 4ac)/4a²]

(x + b/2a) = ± √[(b² - 4ac)]/2a

x = - b/2a ± √[(b² - 4ac)]/2a
x = [- b ± √(b² - 4ac)]/2a ------------------ equation 3.

5) This is the method of the square.

Solution:

(i) 2x² – 7x + 3 = 0

1) The given equation is 2x² – 7x + 3 = 0 ------------------ equation 1.

2) Divide equation 1 by 2 we get,

2x² – 7x + 3 = 0

x² – (7/2)x + (3/2) = 0

x² – (7/2)x = - (3/2) ------------------ equation 2.

3) From equation 1, 

last term = [(middle term)²]/4

last term = [(7/2)²]/4

last term = 49/16

4) Add 49/16 to both sides of equation 2, and we get, 

x² - (7/2)x + 49/16 = 49/16 - 3/2

x² - (7/2)x + 49/16 = 49/16 - 24/16

x² - (7/2)x + 49/16 = (49 - 24)/16
(x - 7/4)² = 25/16

(x - 7/4 = ± √(25/16)

(x - 7/4) = ± 5/4
x = 7/4 ± 5/4
x = (7 ± 5)/4
x = (7 + 5)/4, or x = (7 - 5)/4
x = 12/4, or x = 2/4
x = 3, or x = 1.

5) The roots of the quadratic equation 2x² – 7x + 3 = 0 are 3 and 1.

(ii) 2x² + x – 4 = 0

1) The given equation is 2x² + x - 4 = 0 ------------------ equation 1.

2) Divide equation 1 by 2 we get,

2x² + x - 4 = 0

x² + (1/2)x - (4/2) = 0

x² + (1/2)x = 2 ------------------ equation 2.

3) From equation 1, 

last term = [(middle term)²]/4

last term = [(1/2)²]/4

last term = 1/16

4) Add 1/16 to both sides of equation 2, and we get, 

x² + (1/2)x + 1/16 = 1/16 + 2

x² + (1/2)x + 1/16 = 1/16 + 32/16

x² + (1/2)x + 1/16 = (1 + 32)/16
(x + 1/4)² = 33/16

(x + 1/4) = ± √(33/16)

(x + 1/4) = ± √33/4
x = - (1/4 ± √33/4)
x = (- 1 ± √33)/4
x = (- 1 + √33)/4, or x = (- 1 - √33)/4.

5) The roots of the quadratic equation 2x² + x - 4 = 0 are 

(- 1 + √33)/4 and (- 1 - √33)/4.

(iii) 4x² + 4√3x + 3 = 0

1) The given equation is 4x² + 4√3x + 3 = 0 ------------------ equation 1.

2) Divide equation 1 by 4 we get,

4x² + 4√3x + 3 = 0

x² + √3x + (3/4) = 0

x² + √3x = - (3/4) ------------------ equation 2.

3) From equation 1, 

last term = [(middle term)²]/4

last term = [(√3)²]/4

last term = 3/4

4) Add 3/4 to both sides of equation 2, and we get, 

x² + √3x + (3/4) = (3/4) - (3/4)

x² + √3x + (3/4) = 0

(x + 3/2)² = 0

(x + 3/2) = ± 0

x = - (3/2 ± 0)
x = - 3/2 ± 0
x = - 3/2, or x = - 3/2.

5) The roots of the quadratic equation 4x² + 4√3x + 3 = 0 are -3/2 and -3/2.

(iv) 2x² + x + 4 = 0

1) The given equation is 2x² + x + 4 = 0 ------------------ equation 1.

2) Divide equation 1 by 2 we get,

2x² + x + 4 = 0

x² + (1/2)x + (4/2) = 0

x² + (1/2)x = - 2 ------------------ equation 2.

3) From equation 1, 

last term = [(middle term)²]/4

last term = [(1/2)²]/4

last term = 1/16

4) Add 1/16 to both sides of equation 2, and we get, 

x² + (1/2)x + 1/16 = 1/16 - 2

x² + (1/2)x + 1/16 = 1/16 - 32/16

x² + (1/2)x + 1/16 = (1 - 32)/16
(x + 1/4)² = - 33/16

5) As the square of any real number can't be negative, so 2x² + x + 4 = 0 has no

real roots.

Q2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2x² – 7x + 3 = 0         (ii) 2x² + x – 4 = 0

(iii) 4x² + 4√3x + 3 = 0     (iv) 2x² + x + 4 = 0

Explanation:

1) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

ax² + bx + c = 0

x² + (b/a)x + (c/a) = 0

x² + (b/a)x = - (c/a)
x² + (b/a)x = - (c/a) ------------------ equation 1.

2) In the quadratic equation, for getting the perfect square, 

last term = [(middle term)²]/4 ------------------ equation 2.

3) From equation 1 and equation 2, last term = [(b/a)²]/4.

4) Adding [(b/a)²]/4 to both sides of equation 1, we get,

x² + (b/a)x + [(b/a)²]/4 = [(b/a)²]/4 - (c/a)

x² + (b/a)x + (b/2a)² = (b/2a)² - (c/a)
x² + (b/a)x + (b/2a)² = (b)²/4a² - (c/a)

(x + b/2a)² = (b² - 4ac)/4a²

(x + b/2a) = ± √[(b² - 4ac)/4a²]

(x + b/2a) = ± √[(b² - 4ac)]/2a

x = - b/2a ± √[(b² - 4ac)]/2a
x = [- b ± √(b² - 4ac)]/2a ------------------ equation 3.

5) This is the method of the square.

Solution:

(i) 2x² – 7x + 3 = 0

1) The given equation is 2x² - 7x + 3 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax² + bx + c = 0, where a ≠ 0 can be solved

using the formula:

x = [- b ± √(b² - 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 2x² - 7x + 3 = 0 with ax² + bx + c = 0, we have,

a = 2, b = - 7, c = 3.

4) First we will find:

b² - 4ac = (- 7)² - 4(2)(3)

b² - 4ac = 49 - 24

b² - 4ac = 25 ------------------ equation 3. 

5) As b² - 4ac = 25, it has real roots, so from equation 2 and equation 3, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 7) ± √25]/2(2)
x = (7 ± 5)/4

6) So,  x = (7 + 5)/4 or x = (7 - 5)/4.

x = (12)/4 or x = (2)/4

x = 3 or x = 1/2

7) So, x = 3 or x = 1/2.

(ii) 2x² + x – 4 = 0

1) The given equation is 2x² + x - 4 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax² + bx + c = 0, where a ≠ 0 can be solved

using the formula:

x = [- b ± √(b² - 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 2x² + x - 4 = 0 with ax² + bx + c = 0, we have,

a = 2, b = 1, c = - 4.

4) First we will find:

b² - 4ac = (1)² - 4(2)(- 4)

b² - 4ac = 1 + 32

b² - 4ac = 33 ------------------ equation 3. 

5) As b² - 4ac = 33, it has real roots, so from equation 2 and equation 3, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (1) ± √33]/2(2)
x = (- 1 ± √33)/4

6) So,  x = (- 1 + √33)/4 or x = (- 1 - √33)/4.

(iii) 4x² + 4√3x + 3 = 0

1) The given equation is 4x² + 4√3x + 3 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax² + bx + c = 0, where a ≠ 0 can be solved

using the formula:

x = [- b ± √(b² - 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 4x² + 4√3x + 3 = 0 with ax² + bx + c = 0, so

a = 4, b = 4√3, c = 3.

4) First we will find:

b² - 4ac = (4√3)² - 4(4)(3)

b² - 4ac = 48 - 48

b² - 4ac = 0 ------------------ equation 3. 

5) As, b² - 4ac ≥ 0, it has real roots, so from equation 2 and equation 3, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (1) ± √0]/2(4)
x = (- 1 ± 0)/4

6) So,  x = (- 1)/4 or x = (- 1)/4.

(iv) 2x² + x + 4 = 0

1) The given equation is 2x² + x + 4 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax² + bx + c = 0, where a ≠ 0 can be solved

using the formula:

x = [- b ± √(b² - 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation 2x² + x + 4 = 0 with ax² + bx + c = 0, so

a = 2, b = 1, c = 4.

4) First we will find:

b² - 4ac = (1)² - 4(2)(4)

b² - 4ac = 1 - 32

b² - 4ac = - 31 ------------------ equation 3. 

5) As b² - 4ac = - 31 < 0, so 2x² + x + 4 = 0 has no real roots.

 

Q3. Find the roots of the following equations:

(i) x - (1/x) = 3, x ≠ 0   (ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, x ≠ 7

Explanation:

1) Convert your equation in to the quadratic form ax² + bx + c = 0, where a ≠ 0.

2) Solve this equation to get the values of the variable x.

Solution:

(i) x - (1/x) = 3, x ≠ 0

1) The given equation is 

x - (1/x) = 3

(x² - 1)/x = 3

(x² - 1) = 3x

x² - 3x -1 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax² + bx + c = 0, where a ≠ 0 can be solved

using the formula:

x = [- b ± √(b² - 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation x² - 3x -1 = 0 with ax² + bx + c = 0, so

a = 1, b = - 3, c = 1.

4) First we will find:

b² - 4ac = (- 3)² - 4(1)(1)

b² - 4ac = 9 - 4

b² - 4ac = 5 ------------------ equation 3.

5) As b² - 4ac = 5 > 0, it has real roots, so from equation 2 and equation 3, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (1) ± √0]/2(4)
x = (- 1 ± 0)/4

6) So,  x = (- 1)/4 or x = (- 1)/4.

(ii) 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ - 4, x ≠ 7

1) The given equation is 

1/(x + 4) - 1/(x - 7) = 11/30

[1(x - 7) - 1(x + 4)]/(x + 4)(x - 7) = 11/30

(x - 7 - x - 4)/(x + 4)(x - 7) = 11/30

30(x - 7 - x - 4) = 11 (x + 4)(x - 7)
30(- 11) = 11 (x + 4)(x - 7)
- 30 = (x + 4)(x - 7)
- 30 = x(x - 7) + 4 (x - 7)
- 30 = x² - 7x + 4x - 28
- 30 = x² - 3x - 28
x² - 3x - 28 + 30 = 0
x² - 3x + 2 = 0 ------------------ equation 1.

2) We know that the quadratic equation ax² + bx + c = 0, where a ≠ 0 can be solved

using the formula:

x = [- b ± √(b² - 4ac)]/2a ------------------ equation 2.

3) Equate the coefficient of equation x² - 3x + 2 = 0 with ax² + bx + c = 0, so

a = 1, b = - 3, c = 2.

4) First we will find:

b² - 4ac = (- 3)² - 4(1)(2)

b² - 4ac = 9 - 8

b² - 4ac = 1 ------------------ equation 3.

5) As b² - 4ac = 1 > 0, it has real roots, so from equation 2 and equation 3, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 3) ± √1]/2(1)
x = (3 ± 1)/2

6) So,  x = (4)/2 or x = (2)/2, i.e. x = 2, or x = 1.

Q4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3 Find his present age.

1) Let Rehman's present age be x.

2) So, 3 years ago, his age was (x - 3).

3) 5 years later, his age will be (x + 3).

4) According to the problem,

[1/(x - 3)] + [1/(x + 5)] = 1/3

[1(x + 5) + 1(x - 3)]/[(x + 5)(x - 3)] = 1/3

(x + 5 + x - 3)/(x + 5)(x - 3) = 1/3

3(2x + 2) = (x + 5)(x - 3)
6x + 6 = x² + 5x - 3x - 15
x² + 2x - 15 - 6x - 6 = 0
x² - 4x - 21 = 0 ------------------ equation 1.

5) Equate the coefficient of equation x² - 4x - 21 = 0 with ax² + bx + c = 0, so

a = 1, b = - 4, c = - 21

6) First we will find:

b² - 4ac = (- 4)² - 4(1)(- 21)

b² - 4ac = 16 + 84

b² - 4ac = 100 ------------------ equation 2.

7) As b² - 4ac = 100 > 0, it has real roots, so from equation 1 and equation 2, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 4) ± √100]/2(1)

x = (4 ± 10)/2 

8) So,  x = (14)/2 or x = (- 6)/2, i.e. x = 7, or x = - 3.

9) As age can't be negative, x = 7, so Rehman's present age is 7 years.

Q5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

1) Let Shefali's marks in Mathematics be x.

2) Her English marks will be (30 - x).

3) According to the problem,

(x + 2)(30 - x - 3) = 210

(x + 2)(27 - x) = 210

x(27 - x) + 2(27 - x) = 210

27x - x² + 54 - 2x = 210

27x - 2x - x² + 54 = 210

25x - x² = 210 - 54

25x - x² = 156

x² - 25x + 156 = 0 ------------------ equation 1.

4) Equate the coefficient of equation x² - 25x + 156 = 0 with ax² + bx + c = 0, so

a = 1, b = - 25, c = 156

6) First we will find:

b² - 4ac = (- 25)² - 4(1)(156)

b² - 4ac = 625 - 624

b² - 4ac = 1 ------------------ equation 2.

7) As b² - 4ac = 1 > 0, it has real roots, so from equation 1 and equation 2, we

have,

x = [- b ± √(b² - 4ac)]/2a
x = [- (- 25) ± √1]/2(1)

x = (25 ± 1)/2 

8) So,  x = (26)/2 or x = (24)/2, i.e. x = 13, or x = 12.

9) If x = 13, then 

she got 13 marks in Mathematics and 30 - 13 = 17 marks in English.

10) If x = 12, then 

she got 12 marks in Mathematics and 30 - 12 = 18 marks in English.

Q6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

1) Let the shorter side of a rectangle be x m.

2) So, the longer side of a rectangle will be (x + 30) m 

3) According to the problem,

x² + (x + 30)² = (x + 60)²

x² = (x + 60)² - (x + 30)²              use a² - b² = (a - b)(a + b)

x² = [(x + 60) - (x + 30)] [(x + 60) + (x + 30)]

x² = (x + 60 - x - 30) (x + 60 + x + 30)

x² = 30(2x + 90)

x² = 60x + 2700

x² - 60x - 2700 = 0 ------------------ equation 1.

4) Quadratic equation: x² - 60x - 2700 = 0

here last term is 2700 and its sign is minus, factorise 1 x 2700 in such a way

that their difference will be 60.

1 x 2700 = (30) x (- 90) (30 - 90 = - 60).

x² + 30x - 90x - 2700 = 0

x(x + 30) - 90(x + 30) = 0

(x + 30)(x - 90) = 0

5) So, (x + 30) = 0, or (x - 90) = 0

6) So, x = - 30 or x = 90.

7) As the length of any side can't be negative, ignore x = -30, so we have x = 90.

8) The length of the shorter side is 90 m and the longer side is 90 + 30 = 120 m.

Q7. The difference of the squares of the two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

1) Let the larger number be x.

2) So, the square of the smaller number will be 8x

3) According to the problem,

x² - 8x = 180

x² - 8x - 180 = 0 ------------------ equation 1.

4) Quadratic equation: x² - 8x - 180 = 0 

here last term is 180 and its sign is minus, factorise 1 x 180 in such a way

that their difference will be 8.

1 x 180 = (10) x (- 18) (10 - 18 = - 8).

x² + 10x - 18x - 180 = 0

x(x + 10) - 18(x + 10) = 0

(x + 10)(x - 18) = 0

5) So, (x + 10) = 0, or (x - 18) = 0

6) So, x = - 10 or x = 18.

7) If x = - 10, the smaller number will be - 10 (8) = - 80, as a square of any

number can't be negative, we can't take x = -10 as the lager number.

8) So, our larger number will be 18. 

9) According to the problem, a square of a smaller number = 8x = 8(18).

square of a smaller number = 8(18)

square of a smaller number = 2 x 2 x 2 x 2 x 3 x 3, so,

smaller number = ± (2 x 2 x 3)

smaller number = ± 12.

10) The numbers are 12 and 18 or - 12 and 18.

Q8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

1) Let the uniform speed of a train be x km/h.

2) To cover 360 km, the time will be (360/x) hrs.

3) If the speed is increased by 5 km/h, then time will be decreased by 1 hr, so 

new speed = (x + 5) km/h, and new time = [(360/x) -1] hrs.

4) According to the condition,

(x + 5)[(360/x) -1] = 360

(x + 5)[(360 - x)/x] = 360

(x + 5)(360 - x) = 360x

x(360 - x) + 5(360 - x) = 360x

360x - x² + 1800 - 5x = 360x

- x² + 1800 - 5x = 0

x² + 5x - 1800 = 0 ---------- equation 1

4) Quadratic equation: x² + 5x - 1800 = 0 

here last term is 1800 and its sign is minus, factorise 1 x 1800 in such a way

that their difference will be 5.

1 x 1800 = (45) x (- 40) (45 - 40 = 5).

x² + 45x - 40x - 1800 = 0

x(x + 45) - 40(x + 45) = 0

(x + 45)(x - 40) = 0

5) So, (x + 45) = 0, or (x - 40) = 0

6) So, x = - 45 or x = 40.

7) As x is the speed, it can't be negative, so we have x = 40.

8) The speed of the train is 40 km/h.

Q9.Two water taps together can fill a tank in 9 and 3/8 hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

1) Let the time taken by the smaller pipe to fill the tank be x hrs.

2) Time taken by the larger pipe will be (x - 10) hrs.

3) The part of a tank filled by a smaller pipe in 1 hr will be 1/x.

4) The part of a tank filled by a larger pipe in 1 hr will be 1/(x - 10).

5) Total time taken by both the pipes to fill the tank = 9 and 3/8 = (72+3)/8 = 75/8.

6) According to the problem, the part of a tank filled by both pipes in 1 hour is 8/75.

7) So, we have,

1/x + 1/(x - 10) = 8/75

[(x - 10) + x]/x(x - 10) = 8/75

75(2x - 10) = 8x(x - 10)

150x - 750 = 8x² - 80x

8x² - 80x - 150x + 750 = 0

8x² - 230x + 750 = 0 ---------- equation 1

8) Quadratic equation: 8x² - 230x + 750 = 0 

here last term is 750 and its sign is plus, factorise 8 x 750 in such a way

that their sum will be - 230. 

8 x 750 = (- 200) x (- 30)

(- 200 - 30 = - 230).

8x² - 200x - 30x + 750 = 0

8x(x - 25) - 30(x - 25) = 0

(x - 25)(8x - 30) = 0

9) So, (x - 25) = 0, or (8x - 30) = 0

10) So, x = 25 or x = 30/8.

11) If we consider x = 30/8, the time taken by a larger pipe will be 

(30/8 - 10) = (30 - 80)/8 = - 50/8 which negative, so ignore x = 30/8.

12) So consider x = 25. i.e. time taken by smaller pipe to fill the tank will be 25 hrs

and the time taken by larger pipe to fill the tank will be 15 hrs.

Q10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains. 

1) Let the average speed of a passenger train be x km/h.

2) So the average speed of the express train will be (x + 11) km/h.

3) We know that time = distance/speed, so to cover 132 km,

a) for passenger train, time = 132/x hrs.

b) for express train, time = 132/(x + 11) hrs.

4) As the express train takes 1 hour less than the passenger train to cover 132 km,

so

[132/x]  - [132/(x + 11)] = 1

132[1/x - 1/(x + 11)] = 1

132[(x + 11) - x ]/x(x + 11)] = 1

132(11)/x(x + 11) = 1

132(11) = x(x + 11)

1452 = x² + 11x

x² + 11x - 1452 = 0 ------------ equation 1

5) Quadratic equation: x² + 11x - 1452 = 0 

here last term is 1452 and its sign is minus, factorise 1 x 1452 in such a way

that their difference will be 11. 

1 x 1452 = (44) x (- 33) (44 - 33 = 11).

x² + 44x - 33x - 1452 = 0

x(x + 44) - 33(x + 1452) = 0

(x + 44)(x - 33) = 0

6) So, (x + 44) = 0, or (x - 33) = 0

7) So, x = - 44 or x = 33.

8) As the speed can't be negative, so ignore x = - 44, so the average speed of

passenger train will be 33 km/h and the average speed of the express train will be 44 km/h.

Q11. The sum of the areas of the two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares. 

1) Let the side of the first square be x m.

2) So, its area will be x². 

3) According to the problem, the area of the other square will be (468 - x²). the side

of this square will be √(468 - x²).

4) The perimeter of the first square will be 4x and that of the other will be 

4√(468 - x²) 

5) As the difference between their perimeters is 24,

4x - 4√(468 - x²) = 24

4[x - √(468 - x²)] = 24

x - √(468 - x²) = 6

x - 6 = √(468 - x²)       squaring both side, we get

(x - 6)² = (468 - x²)

x² - 12x + 36 = 468 - x²

x² + x² - 12x + 36 - 468 = 0
2x² - 12x - 432 = 0
x² - 6x - 216 = 0 ------------------ equation 1.

6) Quadratic equation: x² - 6x - 216 = 0

here last term is 216 and its sign is minus, factorise 1 x 216 in such a way

that their difference will be - 6.

1 x 216 = (12) x (- 18) (12 - 18 = - 6).

x² + 12x - 18x - 216 = 0

x(x + 12) - 18(x + 12) = 0

(x + 12)(x - 18) = 0

5) So, (x + 12) = 0, or (x - 18) = 0

6) So, x = - 12 or x = 18.

7) As the length of any side can't be negative, ignore x = - 12, so we have x = 18.

8) The length of the side of the first square will be 18 m.

9) So the Area of the other square is

= 468 - x²

= 468 - 18²
= 468 - 324
= 144. Therefore side of other square will be √144 = 12 m.

10) The sides of the squares are 12 m and 18 m.

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